TITLE: The behavior of quadratic formula in the limit $a\to 0$
QUESTION [7 upvotes]: $$\lim\limits_{a\to0}\frac{-b+\sqrt{b^2-4ac}}{2a}$$
$b$ and $c$ are constants. As $a$ approaches $0$, what does the formula approach?
Example:
$$\lim\limits_{a\to0}\frac{-5+\sqrt{5^2-4·0.001·3}}{2·0.001}\approx-0.6$$

REPLY [9 votes]: $$\begin{align*}
\lim_{a\to 0}\frac{-b+\sqrt{b^2-4ac}}{2a}
&= \lim_{a\to 0}\left(\frac{-b+\sqrt{b^2-4ac}}{2a}\cdot\frac{b+\sqrt{b^2-4ac}}{b+\sqrt{b^2-4ac}}\right)\\
&= \lim_{a\to 0}\frac{-b^2+(b^2-4ac)}{2a(b+\sqrt{b^2-4ac})}\\
&= \lim_{a\to 0}\frac{-2c}{b+\sqrt{b^2-4ac}}\\
&= \frac{-2c}{b+|b|}
\end{align*}$$
If $b>0$, then you get $-\dfrac cb$, the root of the linear equation hinted in another answer; but
if $b<0$, then the limit does not exist, because

the larger one of the two quadratic roots becomes $+\infty$ when $a\to 0^+$;
the smaller one of the two quadratic roots becomes $-\infty$ when $a\to 0^-$;